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What is the antiderivative of #x sin (x/2)#?

Answer 1

Ezra Adams

#int xsin(x/2)dx =-2xcos(x/2) + 4sin(x/2)+C#

Evaluate the indefinite integral:

#int xsin(x/2)dx#

by parts:

#int xsin(x/2)dx = int x d/dx (-2cos(x/2))dx#

#int xsin(x/2)dx =-2xcos(x/2) + 2 int cos(x/2)dx#

#int xsin(x/2)dx =-2xcos(x/2) + 4 int cos(x/2)d(x/2)#

#int xsin(x/2)dx =-2xcos(x/2) + 4sin(x/2)+C#

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Answer 2

Ezra Adams

The antiderivative of (x \sin\left(\frac{x}{2}\right)) is (-2x \cos\left(\frac{x}{2}\right) + 4 \sin\left(\frac{x}{2}\right) + C), where (C) is the constant of integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.